# Solving Trig eqations

• Dec 16th 2007, 11:58 AM
Swim321
Solving Trig eqations
I have two questions:

1-How do I go about solving the equation 3cotē + 5csc =-1
(i couldn't get the angle symbol to appear)

2- I am supposed to solve this eqation (sec(x)=x-3) graphically. I can get the graph just fine but am now confused as to what the answers are.
• Dec 16th 2007, 12:29 PM
wingless
1-
$\displaystyle 3 \text{Cot}^2x + 5 \text{Csc} x = -1$

$\displaystyle 3 \text{Cot}^2x + \frac{5}{\text{Sin} x} + 1 = 0$

$\displaystyle \frac{3\text{Cos}^2x}{\text{Sin}^2x} + \frac{5}{\text{Sin} x} + 1 = 0$

$\displaystyle \frac{3\text{Cos}^2x + 5 \text{Sin} x + \text{Sin}^2x}{\text{Sin}^2x}= 0$

(for $\displaystyle x\neq \text{k$\pi $}$)

$\displaystyle 3\text{Cos}^2x + 5 \text{Sin} x + \text{Sin}^2x= 0$

$\displaystyle 3\left(1 - \text{Sin}^2x\right) + 5 \text{Sin} x + \text{Sin}^2x = 0$

$\displaystyle 3 - 2\text{Sin}^2x + 5 \text{Sin} x = 0$ (multiply both sides by - 1)

$\displaystyle 2\text{Sin}^2x- 5 \text{Sin} x - 3 = 0$

$\displaystyle (2\text{Sin} x + 1)(\text{Sin} x -3) = 0$

$\displaystyle 2\text{Sin} x + 1 =0 < \text{and} > (\text{Sin} x -3) = 0$

As $\displaystyle \text{Sin} x -3$ can' t be equal to $\displaystyle 0$

$\displaystyle 2\text{Sin} x + 1 =0$

$\displaystyle \text{Sin} x =- \frac{1}{2}$

$\displaystyle x = \text{ArcSin} -\frac{1}{2}$

$\displaystyle x = \frac{11\pi }{6} + 2\text{k$\pi $}$

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2- Graph y = Sec(x) and y = x - 3.
x coordinates of the point(s) they intersect are your x's and the y coordinates of these points are your Sec(x) and x-3.
• Dec 16th 2007, 01:44 PM
Soroban
Hello, Swim321!

Quote:

Solve: .$\displaystyle 3\cot^2\!\theta + 5\csc\theta \:=\:-1$

We have: .$\displaystyle 3(\csc^2\!\theta-1) + 5\csc\theta +1\:=\:0 \quad\Rightarrow\quad 3\csc^2\!\theta - 3 + 5\csc\theta + 1 \:=\:0$

We have the quadratic: .$\displaystyle 3\csc^2\!\theta + 5\csc\theta + 3 \:=\:0$

. . which factors: .$\displaystyle (3\csc\theta-1)(\csc x + 2) \:=\:0$

Then we have: .$\displaystyle 3\csc\theta-1\:=\:0\quad\Rightarrow\quad \csc\theta \:=\:\frac{1}{3}$ . . . . which has no real roots

. . and: .$\displaystyle \csc\theta +2\:=\:0\quad\Rightarrow\quad\csc\theta \:=\:-2\quad\Rightarrow\quad \theta \:=\:\begin{Bmatrix}\dfrac{7\pi}{6} + 2\pi n \\ \dfrac{11\pi}{6} + 2\pi n\end{Bmatrix}$